Question

\(\left(\dfrac{x\sqrt{x}-1}{x\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)

a, Rút gọn A

b, Tìm x để A< 0

c, Tìm x nguyên để A có giá trị nguyên

Answer 1

Sửa đề: \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)

a: \(A=\dfrac{x+\sqrt{x}+1-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\cdot\dfrac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)

\(=\dfrac{2\left(\sqrt{x}+1\right)}{2\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

b: Để A<0 thì \(\sqrt{x}-1< 0\)

hay 0<x<1

c: Để A là số nguyên thì \(\sqrt{x}+1-2⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)

hay \(x\in\left\{4;0;9\right\}\)